EXPLANATION OF ALUMINIUM IONIZATIONS
By Prof. L. Kaliambos (Natural philosopher in New Energy) May 13, 2015 Aluminium (or aluminum ) is a chemical element in the boron group with symbol Al and atomic number 13. However unlike for hydrogen, a closed-form solution to the Schrödinger equation for the many-electron atoms like the aluminium atom has not been found. So, under the invalid relativity (EXPERIMENTS REJECT RELATIVITY) various approximations, such as the Hartree–Fock method, could be used to estimate the ground state energies. Under these difficulties I published my paper “ Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008) by analysing carefully the electromagnetic interactions of two spinning electrons of opposite spin. Hence the electronic structure of the aluminium ground state should be given by this correct image with the following electron configuration: 1s2.2s2.2px2.2py2.2pz2.3s2.3px1. According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies ( in eV) are the following: E1 = 5.98577 , E2 = 18.82856 , E3 = 28.44765 , E4 = 119.992 , E5 = 153.825 , E6 = 190.49, E7 = 241.76 , E8 = 284.66, E9 = 33013 , E10 = 398.75, E11 = 442.00 , E12 = 2085.98, and E13 = 2304.141. Here the -( E1 ) gives the binding energy E( 3px1) of the one outer electron. Then the -(E2 + E3) gives the binding energy E(3s2). Also, the - ( E7 + E8 + E9 ) gives the binding energy E(2px1 + 2py1 + 2pz1) of the three electrons with parallel spin , while the -( E4 + E5 + E6 + E7 + E8 + E9 ) gives the binding energy E( 2px2 + 2py2 +2pz2) . On the other hand the -( E10 + E11 ) gives the binding energy E(2s2) , while the -( E12 + E13) gives the binding energy E(1s2). See also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS EXPLANATION OF E1 = -E(3px1) Here the charge (-12e) of the electrons 1s2.2s2.2px2.2py2.2pz2.3ps2 screens the nuclear charge (+13e) and for a perfect screening we would have an effective Zeff = ζ = 1. However the outer (3px1) electron repels the 3ps2 electrons and leads to the deformation of shells with ζ >1. Under this condition I apply the formula of the Bohr model to write the binding energy E(3px1) as E(1px1) = (-13.6057)ζ2/32 = -E1 = -5.98577 Solving for ζ we get ζ = 1.99 >1 ' ' EXPLANATION OF ( E2 + E3 ) = 47.2762 eV = -E(3s2) Here the charges (-10e) of the electrons of 1s2.2s2.2px2.2py2.2pz2 screen the nuclear charge (+13e) and for a perfect screening we would have an effective ζ = 3. However the two electrons of 3s2 penetrate the 2px2.2py2.2pz2 leading to the deformation of shells with ζ > 3. Under this condition I apply my formula of 2008 to write the equation of the binding energy E(3s2) = - (E2 + E3) of the two electrons as ( E2 + E3 ) = 47.2762 eV = - E(3s2) = - 27.21)ζ2 + ( 16.95)ζ - 4.1 /n2 Since n = 3 we may write 3.0233ζ2 - 1.8833ζ - 46.82 = 0 Then solving for ζ we get ζ = 4.26 > 3 EXPLANATION OF -( E7 + E8 + E9 ) = E( 2px1 + 2py1 + 2pz1) ' '''Here the 2px1 , 2py1 , and 2pz1 under a perfect screening of the spherical shells 1s2 and 2s2, should lead to the effective ζ = 9, because +13e -2e -2e = +9e. However the deformation of spherical shells leads to ζ > 9 . Here using the Bohr formula we may write E (2pz1 + 2py1 + 2px1) = 3(-13.6057)ζ2 /22 = - (E7 + E8 + E9 ) = - 856.55 Therefore we get ζ = 9.16 > 9 Here ζ = 9.16 means that the three electrons with parallel spin (S = 1) which exert both electric and magnetic repulsions , exist at symmetrical positions providing an almost perfect screening . ' ' '''EXPLANATION OF' '-(E4 + E5 + E6 + E7 + E8 + E9 ) = -1320.857 = E(2px2 + 2py2 + 2pz2) ' Here we get the total binding energy of 6 electrons by applying my formula of 2008 as E(2px2 + 2py2 + 2pz2) = 3+ (16.95)ζ - 4.1) / 22 = - 1320.857 But solving for ζ we get ζ < 9 , which cannot exist. In fact, since 2px2 , 2py2 , and 2pz2 make a complete spherical shell they provide a perfect screening with ζ = 9. So using ζ = 9 and solving for n we expect to find n > 2 because a perfect screening means that the quantum number n = 2 becomes n > 2. Under this condition for determining here the number n the above equation could be written as E(2px2 + 2py2 + 2pz2) = 3+ (16.95)9 - 4.1) / n2 = - 1320.857 Then solving for n we get n = 2.16 > 2 In other words the three orbitals of paired electrons do not lead to the deformations of 1s2 and 2s2 but differ from the symmetry of (2px1+ 2py1 +2pz1), because they do not exert magnetic repulsions. Here the electric repulsions between the paired electrons of 2px2, 2py2 and 2pz2 make a complete spherical shell and lead to a perfect screening with ζ = 9. Under this condition the simple quantum number n = 2 becomes n = 2.16 . Note that the two electrons of opposite spin (say the 2px2) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy. However in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of the 2px2 or 2py2 or 2pz2 today many physicists believe incorrectly that it is due to the Coulomb repulsion between the two electrons of opposite spin. Under such fallacious ideas I published my paper of 2008. EXPLANATION OF -( E10 + E11 ) = E (2s2) According to the quantum mechanics the two electrons (2s2) penetrate the 1s2 shell. Thus they lead to the deformations of both 1s2 and 2s2 spherical shells giving an effective ζ > 11, because the charge (- 2e) of the two electrons of 1s2 screens the charge (+13e) of nucleus. Since n = 2 we may write ( E10 + E11 ) = - E(2s2) = - )ζ2 (+16.95)ζ - 4.1 / 22 Since ( E10 + E11 ) = 840.75 eV, we may write 6.8025ζ2 - 4.2375ζ - 839.725 = 0 Then, solving for ζ we get ζ = 11.43 > 11. Here ζ = 11.43 > 11 means that the repulsιοns (2s2 - 1s2 ) lead to the deformation of shells, because the two electrons (2s2) or (1s2) of opposite spin behave like one particle. Note that in both cases the repulsions are due to only electric forces of the Coulomb law. Whereas in the case of the three electrons of 2px1, 2py1, and 2pz1 of parallel spin (S = 1) the three electrons interact with both electric and magnetic repulsions from symmetrical positions. EXPLANATION OF -( Ε12 + E13 ) = E( 1s2) ''' As in the case of helium the binding energy E(1s2) is due to the two remaining electrons of 1s2 with n = 1. Thus we expect to calculate the binding energy by applying my formula of 2008 for Z = 13 as E(1s2) = )132 (+ 16.95) 13 - 4.1 /12 = - 4382.24 However the experiments of ionizations give the - (E12 + E13 ) = - 4390.12 . '''In other words one sees here that after the ionizations my formula of 2008 gives the value of 4382.24 eV which is smaller than the experimental value of 4390.12 eV. Under this condition of ionizations I suggest that n = 1 becomes n < 1 due to the fact that the ionizations reduce the electron charges and now the nuclear charge is much greater than the electron charge of the two remaining electrons. So we may write (E12 + E13) = 4390.12 eV = - E(1s2) = - )132 (+16.95)13 - 4.1 / n2 Then solving for n we get n = 0.9991. Category:Fundamental physics concepts